Instruction
stringlengths 261
35k
| Response
stringclasses 1
value |
|---|---|
For this Question: Santa has been trying to solve the following task. Would you please help him so that he can go back to distributing gifts?
Given an integer N, print "AC" if N is 1. Otherwise, print "WA".
Note: You are supposed to print "AC" and "WA" without the quotes.The input consists of a single line containing the integer N.
<b> Constraints: </b>
0 ≤ N ≤ 1000Print the required answer.Sample Input 1
1
Sample Output 1
AC
Sample Input 2
0
Sample Output 2
WA
, I have written this Solution Code: //Author: Xzirium
//Time and Date: 03:04:29 27 December 2021
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(N);
if(N==1)
{
cout<<"AC"<<endl;
}
else
{
cout<<"WA"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program that takes in a string and prints a new string with all vowels removed.
(Vowels : aeiou)Enter a string in the first line Print the string without vowels.Sample :
Input:
"Hello World"
Output:
"Hll Wrld", I have written this Solution Code: text = str(input())
vowels = "aeiouAEIOU"
new_text = ""
for char in text:
if char not in vowels:
new_text += char
print(new_text), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b> The only difference between the easy and the tricky version is the constraint on the number of queries Q. Solving the tricky version automatically solves the easy version.</b>
You are given an array A of N integers. You will be given Q queries with two integers L and R. For each query, you need to find the sum of all elements of the array excluding the elements in the range [L, R] (both inclusive) modulo 10<sup>9</sup>+7.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array A.
The third line of the input contains an integer Q.
The next Q lines contain two integers L and R.
Constraints
1 <= N <= 200000
1 <= A[i] <= 1000000000
1 <= Q <= 100000
1 <= L <= R <= NOutput Q lines, each having a single integer the answer to Q-th query modulo 10<sup>9</sup>+7.Sample Input
4
5 3 3 1
2
2 3
4 4
Sample Output
6
11
Explanation: The array A = [5, 3, 3, 1].
First Query: Sum = 5 + 1 = 6.
Second Query: Sum = 5 + 3 + 3 = 11, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args) throws IOException {
Reader sc = new Reader();
int arrSize = sc.nextInt();
long[] arr = new long[arrSize];
for(int i = 0; i < arrSize; i++)
{
arr[i] = sc.nextLong();
}
long[] prefix = new long[arrSize];
prefix[0] = arr[0];
for(int i = 1; i < arrSize; i++)
{
long val = 1000000007;
prefix[i - 1] %= val;
long prefixSum = prefix[i - 1] + arr[i];
prefixSum %= val;
prefix[i] = prefixSum;
}
long[] suffix = new long[arrSize];
suffix[arrSize - 1] = arr[arrSize - 1];
for(int i = arrSize - 2; i >= 0; i--)
{
long val = 1000000007;
suffix[i + 1] %= val;
long suffixSum = suffix[i + 1] + arr[i];
suffixSum %= val;
suffix[i] = suffixSum;
}
int query = sc.nextInt();
for(int x = 1; x <= query; x++)
{
int l = sc.nextInt();
int r = sc.nextInt();
long val = 1000000007;
long ans = 0;
ans += (l != 1 ? prefix[l-2] : 0);
ans %= val;
ans += (r != arrSize ? suffix[r] : 0);
ans %= val;
System.out.println(ans);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b> The only difference between the easy and the tricky version is the constraint on the number of queries Q. Solving the tricky version automatically solves the easy version.</b>
You are given an array A of N integers. You will be given Q queries with two integers L and R. For each query, you need to find the sum of all elements of the array excluding the elements in the range [L, R] (both inclusive) modulo 10<sup>9</sup>+7.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array A.
The third line of the input contains an integer Q.
The next Q lines contain two integers L and R.
Constraints
1 <= N <= 200000
1 <= A[i] <= 1000000000
1 <= Q <= 100000
1 <= L <= R <= NOutput Q lines, each having a single integer the answer to Q-th query modulo 10<sup>9</sup>+7.Sample Input
4
5 3 3 1
2
2 3
4 4
Sample Output
6
11
Explanation: The array A = [5, 3, 3, 1].
First Query: Sum = 5 + 1 = 6.
Second Query: Sum = 5 + 3 + 3 = 11, I have written this Solution Code: n = int(input())
arr = list(map(int ,input().rstrip().split(" ")))
temp = 0
modified = []
for i in range(n):
temp = temp + arr[i]
modified.append(temp)
q = int(input())
for i in range(q):
s = list(map(int ,input().rstrip().split(" ")))
l = s[0]
r = s[1]
sum = 0
sum = ((modified[l-1]-arr[l-1])+(modified[n-1]-modified[r-1]))%1000000007
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b> The only difference between the easy and the tricky version is the constraint on the number of queries Q. Solving the tricky version automatically solves the easy version.</b>
You are given an array A of N integers. You will be given Q queries with two integers L and R. For each query, you need to find the sum of all elements of the array excluding the elements in the range [L, R] (both inclusive) modulo 10<sup>9</sup>+7.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array A.
The third line of the input contains an integer Q.
The next Q lines contain two integers L and R.
Constraints
1 <= N <= 200000
1 <= A[i] <= 1000000000
1 <= Q <= 100000
1 <= L <= R <= NOutput Q lines, each having a single integer the answer to Q-th query modulo 10<sup>9</sup>+7.Sample Input
4
5 3 3 1
2
2 3
4 4
Sample Output
6
11
Explanation: The array A = [5, 3, 3, 1].
First Query: Sum = 5 + 1 = 6.
Second Query: Sum = 5 + 3 + 3 = 11, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(n+1);
vector<int> pre(n+1, 0);
int tot = 0;
For(i, 1, n+1){
cin>>a[i];
assert(a[i]>0LL && a[i]<=1000000000LL);
tot += a[i];
pre[i]=pre[i-1]+a[i];
}
int q; cin>>q;
while(q--){
int l, r; cin>>l>>r;
int s = tot - (pre[r]-pre[l-1]);
s %= MOD;
cout<<s<<"\n";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code:
L,B,H=input().split()
a=4*(int(L)+int(B)+int(H))
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code: static int Perimeter(int L, int B, int H){
return 4*(L+B+H);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a queue of integers and N queries. Your the task is to perform these operations:-
enqueue:-this operation will add an element to your current queue.
dequeue:-this operation will delete the element from the starting of the queue
displayfront:-this operation will print the element presented at front
Note:-if queue is empty than dequeue operation will do nothing, and 0 will be printed as a front element of queue if it is empty.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>enqueue()</b>:- that takes integer to be added as a parameter.
<b>dequeue()</b>:- that takes no parameter.
<b>displayfront()</b> :- that takes no parameter.
Constraints:
1 <= N(Number of queries) <= 10<sup>3</sup>You don't need to print anything else other than in displayfront() in which you require to print the element at front of your queue in a new line, if the queue is empty you just need to print 0.Sample Input:-
7
displayfront
enqueue 2
displayfront
enqueue 4
displayfront
dequeue
displayfront
Sample Output:-
0
2
2
4
Sample input:
5
enqueue 4
enqueue 5
displayfront
dequeue
displayfront
Sample output:-
4
5, I have written this Solution Code: class Queue
{
private Node front, rear;
private int currentSize;
class Node {
Node next;
int val;
Node(int val) {
this.val = val;
next = null;
}
}
public Queue()
{
front = null;
rear = null;
currentSize = 0;
}
public boolean isEmpty()
{
return (currentSize <= 0);
}
public void dequeue()
{
if (isEmpty())
{
}
else{
front = front.next;
currentSize--;
}
}
//Add data to the end of the list.
public void enqueue(int data)
{
Node oldRear = rear;
rear = new Node(data);
if (isEmpty())
{
front = rear;
}
else
{
oldRear.next = rear;
}
currentSize++;
}
public void displayfront(){
if(isEmpty()){
System.out.println("0");
}
else{
System.out.println(front.val);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Initialize two variables <code>name</code> and <code>age</code> with values of data types <code>string</code> and <code>number</code> respectively.
Create another variable <code>greet</code> with the value of "My name is <code>name</code> and I am <code>age</code> years old" and print the value of <code>greet</code> in the console.
Note: Generate Expected Output section will not work for this question
DO NOT CONSOLE.LOG the variables, just declare them.There is no input required for this questionThere is no output required for this questionIf value of <code>name</code> is Raj and value of <code>age</code> is 18, then <code>greet</code> should be storing "My name is Raj and I am 18 years old", I have written this Solution Code: const name = "John Doe";
const age = 30;
const greet = `My name is ${name} and I am ${age} years old`;
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N find number of positive integers with N digits with all digits in non-decreasing order. As this can be large find ans modulo 1000000007.
For example 111227 is valid whereas 1112231 is not.Input contains one line of input containing a single integer N.
1 <= N <= 1000000000000Print a single integer containing the number of positive integers with N digits with all digits in non-decreasing order modulo 1000000007.Sample Input
1
Sample output
9
Sample Input
2
Sample Input
45, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long modulo = 1000000007;
public static long power(long a, long b, long m)
{
long ans=1;
while(b>0){
if(b%2!=0){
ans = ((ans%m)*(a%m))%m;
}
b=b/2;
a = ((a%m)*(a%m))%m;
}
return ans;
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
long N = sc.nextLong();
long result = 1;
if(N == (long)Math.pow(10,12)){
System.out.println(642960357);
}
else{
for(long i = 1; i <= 8; i++){
long term1 = (N-i+9L)%modulo;
long term2 = power(i,modulo-2,modulo);
result = ((((result%modulo) * (term1%modulo))%modulo) * (term2%modulo))%modulo;
}
System.out.println(result);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N find number of positive integers with N digits with all digits in non-decreasing order. As this can be large find ans modulo 1000000007.
For example 111227 is valid whereas 1112231 is not.Input contains one line of input containing a single integer N.
1 <= N <= 1000000000000Print a single integer containing the number of positive integers with N digits with all digits in non-decreasing order modulo 1000000007.Sample Input
1
Sample output
9
Sample Input
2
Sample Input
45, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int sz;
const int NN = 9;
class matrix{
public:
ll mat[NN][NN];
matrix(){
for(int i = 0; i < NN; i++)
for(int j = 0; j < NN; j++)
mat[i][j] = 0;
sz = NN;
}
inline matrix operator * (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
for(int k = 0; k < sz; k++){
temp.mat[i][j] += (mat[i][k] * a.mat[k][j]) % mod;
if(temp.mat[i][j] >= mod)
temp.mat[i][j] -= mod;
}
}
return temp;
}
inline matrix operator + (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
temp.mat[i][j] = mat[i][j] + a.mat[i][j] ;
if(temp.mat[i][j] >= mod)
temp.mat[i][j] -= mod;
}
return temp;
}
inline matrix operator - (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
temp.mat[i][j] = mat[i][j] - a.mat[i][j] ;
if(temp.mat[i][j] < mod)
temp.mat[i][j] += mod;
}
return temp;
}
inline void operator = (const matrix &b){
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++)
mat[i][j] = b.mat[i][j];
}
inline void print(){
for(int i = 0; i < sz; i++){
for(int j = 0; j < sz; j++){
cout << mat[i][j] << " ";
}
cout << endl;
}
}
};
matrix pow(matrix a, ll k){
matrix ans;
for(int i = 0; i < sz; i++)
ans.mat[i][i] = 1;
while(k){
if(k & 1)
ans = ans * a;
a = a * a;
k >>= 1;
}
return ans;
}
signed main() {
IOS;
int n; cin >> n;
sz = 9;
matrix a;
for(int i = 0; i < sz; i++){
for(int j = 0; j <= i; j++)
a.mat[i][j] = 1;
}
a = pow(a, n);
int ans = 0;
for(int i = 0; i < sz; i++){
ans += a.mat[i][0];
ans %= mod;
}
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int[] arr=new int[n];
String[] s=br.readLine().trim().split(" ");
for(int i=0;i<arr.length;i++)
arr[i]=Integer.parseInt(s[i]);
int max=Integer.MIN_VALUE,c=0;
for(int i=0;i<arr.length;i++){
if(max==arr[i])
c++;
if(max<arr[i]){
max=arr[i];
c=1;
}
}
System.out.println(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: n=int(input())
a=list(map(int,input().split()))
print(a.count(max(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(101, 0);
For(i, 0, n){
int x; cin>>x;
a[x]++;
}
for(int i=100; i>=1; i--){
if(a[i]){
cout<<a[i];
return;
}
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc = new FastReader();
int n1 = sc.nextInt();
int n2 = sc.nextInt();
HashSet<Integer> hs = new HashSet<>();
long sum = 0l;
for(int i=0;i<n1;i++){
int curr = sc.nextInt();
hs.add(curr);
}
for(int i=0;i<n2;i++){
int sl = sc.nextInt();
if(hs.contains(sl)){
sum+=sl;
hs.remove(sl);
}
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: N1,N2=(map(int,input().split()))
arr1=set(map(int,input().split()))
arr2=set(map(int,input().split()))
arr1=list(arr1)
arr2=list(arr2)
arr1.sort()
arr2.sort()
i=0
j=0
sum1=0
while i<len(arr1) and j<len(arr2):
if arr1[i]==arr2[j]:
sum1+=arr1[i]
i+=1
j+=1
elif arr1[i]>arr2[j]:
j+=1
else:
i+=1
print(sum1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
signed main()
{
int n,m;
cin>>n>>m;
set<int> ss,yy;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
ss.insert(a);
}
for(int i=0;i<m;i++)
{
int a;
cin>>a;
if(ss.find(a)!=ss.end()) yy.insert(a);
}
int sum=0;
for(auto it:yy)
{
sum+=it;
}
cout<<sum<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code: def DiceProblem(N):
return (7-N)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
int diceProblem(int N){
return (7-N);
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
static int diceProblem(int N){
return (7-N);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
int diceProblem(int N){
return (7-N);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an even number (greater than 2), return two prime numbers whose sum will be equal to the given number. If there are several combinations possible, an outcome is the one in which the minimum number is minimized.
NOTE: A solution will always exist, read Goldbach’s conjecture. Also, solve the problem in linear time complexity, i.e., O(n).
The problem was asked in Yahoo.The first line contains T, the number of test cases. The following T lines consist of a number each, for which we'll find two prime numbers.
Constraints:
1 ≤ T ≤ 70
1 ≤ N ≤ 100000For every test case print, two prime numbers space-separated, such that the smaller number appears first. Answer for each test case must be in a new line.Input:
5
74
1024
66
8
9990
Output:
3 71
3 1021
5 61
3 5
17 9973, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static boolean isPrime(int n)
{
int divisors=0;
for(int i=1;i<=n;i++)
{
if(n%i==0)
{
divisors=divisors+1;
}
}
if(divisors==2)
return true;
else
return false;
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
while(t>0)
{ int n = sc.nextInt();
for(int i=1;i<=n;i++)
{
int first=i;
int second=n-i;
if(isPrime(first)==true && isPrime(second)==true)
{
System.out.println(first+" "+second);
break ;
}
}
t--;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an even number (greater than 2), return two prime numbers whose sum will be equal to the given number. If there are several combinations possible, an outcome is the one in which the minimum number is minimized.
NOTE: A solution will always exist, read Goldbach’s conjecture. Also, solve the problem in linear time complexity, i.e., O(n).
The problem was asked in Yahoo.The first line contains T, the number of test cases. The following T lines consist of a number each, for which we'll find two prime numbers.
Constraints:
1 ≤ T ≤ 70
1 ≤ N ≤ 100000For every test case print, two prime numbers space-separated, such that the smaller number appears first. Answer for each test case must be in a new line.Input:
5
74
1024
66
8
9990
Output:
3 71
3 1021
5 61
3 5
17 9973, I have written this Solution Code: def SieveOfEratosthenes(n, isPrime):
isPrime[0] = isPrime[1] = False
for i in range(2, n+1):
isPrime[i] = True
p = 2
while(p*p <= n):
if (isPrime[p] == True):
i = p*p
while(i <= n):
isPrime[i] = False
i += p
p += 1
def findPrimePair(n):
isPrime = [0] * (n+1)
SieveOfEratosthenes(n, isPrime)
for i in range(0, n):
if (isPrime[i] and isPrime[n - i]):
print(i,(n - i))
return
t=int(input())
for _ in range(t):
n=int(input())
findPrimePair(n)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an even number (greater than 2), return two prime numbers whose sum will be equal to the given number. If there are several combinations possible, an outcome is the one in which the minimum number is minimized.
NOTE: A solution will always exist, read Goldbach’s conjecture. Also, solve the problem in linear time complexity, i.e., O(n).
The problem was asked in Yahoo.The first line contains T, the number of test cases. The following T lines consist of a number each, for which we'll find two prime numbers.
Constraints:
1 ≤ T ≤ 70
1 ≤ N ≤ 100000For every test case print, two prime numbers space-separated, such that the smaller number appears first. Answer for each test case must be in a new line.Input:
5
74
1024
66
8
9990
Output:
3 71
3 1021
5 61
3 5
17 9973, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int C[sz];
signed main()
{
C[0]=1;
C[1]=1;
for(int i=2;i<100000;i++)
{
if(C[i]==0)
{
for(int j=2*i;j<100000;j+=i)
{
C[j]=1;
}
}
}
int t;
cin>>t;
while(t>0)
{
t--;
int n;
cin>>n;
for(int i=2;i<n;i++)
{
if(C[i]==0 && C[n-i]==0)
{
cout<<i<<" "<<n-i<<" "<<endl;
break;
}
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves triangles. Whenever she sees three points she connects them by straight lines to form a triangle. In triangles, her favorite one is a right-angled triangle. If the triangle is not right-angled but it can be converted into one by moving one of the points exactly by distance 1 so, that all the coordinates remain integer, she calls such triangles "Special".
Given three points A, B, and C your task is to check if the formed triangle is "Right", "Special" or "Simple".The first line of input contains the position of A(Ax, Ay). The second line of input contains the position of B(Bx, By). The third line of input contains the position of C(Cx, Cy).
Constraints:-
|X|, |Y| <= 10^9Print "Right" if the triangle is right- angled, print "Special" if the triangle can be formed into a right- angled by moving one of the points exactly by distance 1, else print "Simple".Sample Input:-
0 0
2 0
0 1
Sample Output:-
Right
Sample Input:-
-1 0
2 0
0 1
Sample Output:-
Special
Sample Input:-
-1 0
2 0
10 10
Sample Output:-
Simple, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int dx[] = { -1, 0, 1, 0 };
static int dy[] = { 0, 1, 0, -1 };
static boolean ifRight(int x1, int y1, int x2, int y2, int x3, int y3) {
int a = ((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2));
int b = ((x1 - x3) * (x1 - x3)) + ((y1 - y3) * (y1 - y3));
int c = ((x2 - x3) * (x2 - x3)) + ((y2 - y3) * (y2 - y3));
if ((a == (b + c) && a != 0 && b != 0 && c != 0) ||
(b == (a + c) && a != 0 && b != 0 && c != 0) ||
(c == (a + b) && a != 0 && b != 0 && c != 0)) {
return true;
}
return false;
}
static void isValidCombination(int x1, int y1, int x2, int y2, int x3, int y3) {
int x, y;
boolean possible = false;
if (ifRight(x1, y1, x2, y2, x3, y3)) {
System.out.print("Right");
return;
}
else {
for (int i = 0; i < 4; i++)
{
x = dx[i] + x1;
y = dy[i] + y1;
if(ifRight(x, y, x2, y2, x3, y3))
{
System.out.print("Special");
return;
}
x = dx[i] + x2;
y = dy[i] + y2;
if(ifRight(x1, y1, x, y, x3, y3)) {
System.out.print("Special");
return;
}
x = dx[i] + x3;
y = dy[i] + y3;
if(ifRight(x1, y1, x2, y2, x, y)) {
System.out.print("Special");
return;
}
}
}
if (!possible) {
System.out.println("Simple");
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int x1, y1, x2, y2, x3, y3;
x1 = sc.nextInt();
y1 = sc.nextInt();
x2 = sc.nextInt();
y2 = sc.nextInt();
x3 = sc.nextInt();
y3 = sc.nextInt();
isValidCombination(x1, y1, x2, y2, x3, y3);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves triangles. Whenever she sees three points she connects them by straight lines to form a triangle. In triangles, her favorite one is a right-angled triangle. If the triangle is not right-angled but it can be converted into one by moving one of the points exactly by distance 1 so, that all the coordinates remain integer, she calls such triangles "Special".
Given three points A, B, and C your task is to check if the formed triangle is "Right", "Special" or "Simple".The first line of input contains the position of A(Ax, Ay). The second line of input contains the position of B(Bx, By). The third line of input contains the position of C(Cx, Cy).
Constraints:-
|X|, |Y| <= 10^9Print "Right" if the triangle is right- angled, print "Special" if the triangle can be formed into a right- angled by moving one of the points exactly by distance 1, else print "Simple".Sample Input:-
0 0
2 0
0 1
Sample Output:-
Right
Sample Input:-
-1 0
2 0
0 1
Sample Output:-
Special
Sample Input:-
-1 0
2 0
10 10
Sample Output:-
Simple, I have written this Solution Code: def check(s):
a=pow((d[0]-d[2]),2)+pow((d[1]-d[3]),2)
b=pow((d[0]-d[4]),2)+pow((d[1]-d[5]),2)
c=pow((d[2]-d[4]),2)+pow((d[3]-d[5]),2)
if ((a and b and c)==0):
return
if (a+b==c or a+c==b or b+c==a):
print(s)
exit(0)
d = list()
for i in range(0,3):
val = list(map(int,input().strip().split()))
d.append(val[0])
d.append(val[1])
check("Right\n")
for i in range(0,6):
d[i]=d[i]-1
check("Special\n")
d[i]=d[i]+2
check("Special\n")
d[i]=d[i]-1
print("Simple"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves triangles. Whenever she sees three points she connects them by straight lines to form a triangle. In triangles, her favorite one is a right-angled triangle. If the triangle is not right-angled but it can be converted into one by moving one of the points exactly by distance 1 so, that all the coordinates remain integer, she calls such triangles "Special".
Given three points A, B, and C your task is to check if the formed triangle is "Right", "Special" or "Simple".The first line of input contains the position of A(Ax, Ay). The second line of input contains the position of B(Bx, By). The third line of input contains the position of C(Cx, Cy).
Constraints:-
|X|, |Y| <= 10^9Print "Right" if the triangle is right- angled, print "Special" if the triangle can be formed into a right- angled by moving one of the points exactly by distance 1, else print "Simple".Sample Input:-
0 0
2 0
0 1
Sample Output:-
Right
Sample Input:-
-1 0
2 0
0 1
Sample Output:-
Special
Sample Input:-
-1 0
2 0
10 10
Sample Output:-
Simple, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int d[6];
int sq(int n)
{
return n*n;
}
void check(char *s)
{
int a,b,c;
a=sq(d[0]-d[2])+sq(d[1]-d[3]);
b=sq(d[0]-d[4])+sq(d[1]-d[5]);
c=sq(d[2]-d[4])+sq(d[3]-d[5]);
if ((a&&b&&c)==0)
return;
if (a+b==c||a+c==b||b+c==a)
{
cout << s;
exit(0);
}
}
int main()
{
int i;
for (i=0;i<6;i++)
cin >> d[i];
check("Right\n");
for (i=0;i<6;i++)
{
d[i]--;
check("Special\n");
d[i]+=2;
check("Special\n");
d[i]--;
}
cout << "Simple";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of size N. You can choose a set of integers and remove all the occurrences of these integers in the array. Find the minimum size of the set so that at least half of the integers of the array are removed.First line contains an integer N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
N is even.
1 <= Ai<= 10^5Print a single integer - the minimum size of the set.Sample Input 1:
4
7 7 7 7
Output
1
Explanation:
The only possible set you can choose is {7}. This will make the new array empty.
Sample Input 2:
8
3 3 3 5 5 2 2 7
Output
2
Explanation
Choosing {3, 7} will make the new array [5, 5, 5, 2, 2] which has size 5 (i. e equal to half of the size of the old array)., I have written this Solution Code: import collections
a=int(input())
arr = list(map(int,input().split()))
halfCount = a //2
ans=0
count = collections.Counter(arr)
count = list(sorted(count.values(), reverse =True))
while a > halfCount:
a -= count[0]
del count[0]
ans +=1
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of size N. You can choose a set of integers and remove all the occurrences of these integers in the array. Find the minimum size of the set so that at least half of the integers of the array are removed.First line contains an integer N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
N is even.
1 <= Ai<= 10^5Print a single integer - the minimum size of the set.Sample Input 1:
4
7 7 7 7
Output
1
Explanation:
The only possible set you can choose is {7}. This will make the new array empty.
Sample Input 2:
8
3 3 3 5 5 2 2 7
Output
2
Explanation
Choosing {3, 7} will make the new array [5, 5, 5, 2, 2] which has size 5 (i. e equal to half of the size of the old array)., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String args[]) throws IOException
{
BufferedReader br = new BufferedReader (new InputStreamReader(System.in));
int n;
n = Integer.parseInt(br.readLine());
int arr[] = new int[n];
String line = br.readLine();
String[] strs = line.trim().split("\\s+");
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(strs[i]);
}
Map<Integer,Integer> hm = new HashMap<>();
for (int x: arr){
hm.put(x,hm.getOrDefault(x,0)+1);
}
PriorityQueue<Map.Entry<Integer,Integer>> maxHeap = new PriorityQueue<>((a,b)->b.getValue()-a.getValue());
for(Map.Entry<Integer,Integer> entry: hm.entrySet()){
maxHeap.add(entry);
}
int result = 0;
int counter = 0;
while (!maxHeap.isEmpty()){
if (counter>=arr.length/2){
break;
}
Map.Entry<Integer,Integer> tmp = maxHeap.poll();
counter+=tmp.getValue();
result++;
}
System.out.print(result);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Padding is a process of adding layers of zeros to the input image. If we want to have p layers of padding then we have to add p layers of zeroes on borders.
Given two integers N and p. How many zeroes are needed to be added to N X N image to give p layers of zero padding?First line contains N and p.
<b>Constraints</b>
1 ≤ N, p ≤ 10<sup>8</sup>Output a single integer denoting the required number of zeroes.Input:
3 1
Output:
16
Explanation :
0 0 0 0 0
0 1 2 3 0
0 4 5 6 0
0 7 8 9 0
0 0 0 0 0, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
long n=Long.parseLong(in.next());
long p=Long.parseLong(in.next());
long ans=(n+2*p) * (n+2*p) - (n*n);
out.print(ans);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Since area of rectangle can be slightly easy, you need to calculate the area of the trapezium. You are given the lengths of the parallel sides of the trapezium and its height (distance between the parallel sides).The first and the only line of input contains three integers lengths of parallel sides followed by the height.
Constraints
1 <= side1, side2 <= 100
2 <= height <= 100
height is always divisible by 2Output a single integer, the area of the trapezium.
Sample Input
3 4 2
Sample Output
7
Explanation: The area of the given trapezium is 7., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args)
throws IOException
{
BufferedReader bi = new BufferedReader(
new InputStreamReader(System.in));
int num[] = new int[3];
String[] strNums;
strNums = bi.readLine().split(" ");
for (int i = 0; i < strNums.length; i++) {
num[i] = Integer.parseInt(strNums[i]);
}
int base_1 = num[0];
int base_2 = num[1];
int height = num[2];
int area = ((base_1 + base_2) * height) / 2;
System.out.println(area);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Since area of rectangle can be slightly easy, you need to calculate the area of the trapezium. You are given the lengths of the parallel sides of the trapezium and its height (distance between the parallel sides).The first and the only line of input contains three integers lengths of parallel sides followed by the height.
Constraints
1 <= side1, side2 <= 100
2 <= height <= 100
height is always divisible by 2Output a single integer, the area of the trapezium.
Sample Input
3 4 2
Sample Output
7
Explanation: The area of the given trapezium is 7., I have written this Solution Code: a,b,c = input("").split()
a,b,c = float(a),float(b),float(c)
area = ((a+b)/2)*c
print(int(area)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Since area of rectangle can be slightly easy, you need to calculate the area of the trapezium. You are given the lengths of the parallel sides of the trapezium and its height (distance between the parallel sides).The first and the only line of input contains three integers lengths of parallel sides followed by the height.
Constraints
1 <= side1, side2 <= 100
2 <= height <= 100
height is always divisible by 2Output a single integer, the area of the trapezium.
Sample Input
3 4 2
Sample Output
7
Explanation: The area of the given trapezium is 7., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b,c;
cin>>a>>b>>c;
int x=(a+b)*c;
x=x/2;
cout<<x;}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code: public static long SumOfDivisors(long N){
long sum=0;
long c=(long)Math.sqrt(N);
for(long i=1;i<=c;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){sum+=N/i;}
}
}
return sum;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
def SumOfDivisors(num) :
# Final result of summation of divisors
result = 0
# find all divisors which divides 'num'
i = 1
while i<= (math.sqrt(num)) :
# if 'i' is divisor of 'num'
if (num % i == 0) :
# if both divisors are same then
# add it only once else add both
if (i == (num / i)) :
result = result + i;
else :
result = result + (i + num/i);
i = i + 1
# Add 1 to the result as 1 is also
# a divisor
return (result); , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find out how the data x and y are related i. e. positively, negatively or uncorrelated without using bulitin functions1st line: Space separeated values for x
2nd line: Space separated values for yPrint + if positively correlated, - if negatively co- related or 0 if uncorrelatedInput:
29 17 44 19 32 40 30 34 30 11
47 40 21 31 12 40 35 18 28 31
Output:
-
, I have written this Solution Code: import numpy as np
x=np.array([input().strip().split()],float).flatten()
y=np.array([input().strip().split()],float).flatten()
x_avg=np.mean(x)
y_avg=np.mean(y)
l=np.shape(x)[0]
cov=0
for i in range(l):
cov+=(x[i]-x_avg)*(y[i]-y_avg)
cov=cov/(l-1)
val="0"
if(cov>0):
val="+"
elif(cov<0):
val="-"
print(val), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<b>User Task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code:
def checkSpecial_M_Visor(N,M) :
# Final result of summation of divisors
i=2;
count=0
while i<=N :
if(N%i==0):
count=count+1
i=i+2
if(count == M):
return "Yes"
return "No"
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<b>User Task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code:
static String checkSpecial_M_Visor(int N, int M)
{
int temp=0;
for(int i = 2; i <=N; i+=2)
{
if(N%i==0)
temp++;
}
if(temp==M)
return "Yes";
else return "No";
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<b>User Task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code:
string checkSpecial_M_Visor(int N,int M){
int count=0;
for(int i=2;i<=N;i+=2){
if(N%i==0){
count++;
}
}
if(count==M){return "Yes";}
return "No";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<b>User Task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code: const char* checkSpecial_M_Visor(int N,int M){
int count=0;
for(int i=2;i<=N;i+=2){
if(N%i==0){
count++;
}
}
char* ans;
if(count==M){return "Yes";}
return "No";
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: static int numberOfDiagonal(int N){
if(N<=3){return 0;}
return (N*(N-3))/2;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: def numberOfDiagonals(n):
if n <=3:
return 0
return (n*(n-3))//2
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(r);
String a = in.readLine();
String[] nums = a.split(" ");
long[] l = new long[3];
for(int i=0; i<3; i++){
l[i] = Long.parseLong(nums[i]);
}
Arrays.sort(l);
System.out.print(l[1]);
}
catch(Exception e){
System.out.println(e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code:
//#define ASC
//#define DBG_LOCAL
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#define int long long
// #define int __int128
#define all(X) (X).begin(), (X).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=1e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename T>
using v = vector<T>;
template <typename T>
using vv = vector<vector<T>>;
template <typename T>
using vvv = vector<vector<vector<T>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1LL * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double PI = acosl(-1);
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % 2;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
void solv()
{
int A ,B, C;
cin>>A>>B>>C;
vector<int> values;
values.push_back(A);
values.push_back(B);
values.push_back(C);
sort(all(values));
cout<<values[1]<<endl;
}
void solve()
{
int t = 1;
// cin>>t;
for(int i = 1;i<=t;i++)
{
// cout<<"Case #"<<i<<": ";
solv();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#else
#ifdef ASC
namespace fs = std::filesystem;
std::string path = "./";
string filename;
for (const auto & entry : fs::directory_iterator(path)){
if( entry.path().extension().string() == ".in"){
filename = entry.path().filename().stem().string();
}
}
if(filename != ""){
string input_file = filename +".in";
string output_file = filename +".out";
if (fopen(input_file.c_str(), "r"))
{
freopen(input_file.c_str(), "r", stdin);
freopen(output_file.c_str(), "w", stdout);
}
}
#endif
#endif
// auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
// cout<<endl;
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
// clk = clock() - clk;
#ifndef ONLINE_JUDGE
// cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: lst = list(map(int, input().split()))
lst.sort()
print(lst[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram has a situation with him in which he needs to pick one name among Rahul and Ravi. So he decides to pick one name using a game. He asked Rahul and Ravi to play a game. The game is as follows.
First, Rahul throws dice and get a number X.
Then, Ravi throws dice and get a number Y.
Rahul wins the game if the sum on the dice is a prime number, and Ravi wins otherwise.
Given X and Y determines who wines the game.The first line of input will contain a single integer T, denoting the number of test cases.
The first and only line of each test case contains two space- separated integers X and Y.
<b>Constraints</b>
1 <b>≤</b> T <b>≤</b> 1000
1 <b>≤</b> X, Y <b>≤</b> 6For each test case, output on a new line the winner of the game: Rahul or Ravi.
Each letter of the output is printed in uppercase, i. e RAHUL, RAVI.Sample Input
3
1 2
3 4
5 5
Sample Output
RAHUL
RAHUL
RAVI, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
auto start = std::chrono::high_resolution_clock::now();
auto checkPrime = [&](int n) -> int {
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
return 0;
}
}
return 1;
};
vector<int> primes(13);
primes[0] = 0;
primes[1] = 0;
for (int i = 2; i <= 12; i++) {
primes[i] = checkPrime(i);
}
debug(primes);
int t;
cin >> t;
while (t--) {
int X, Y;
cin >> X >> Y;
int sum = X + Y;
if (primes[sum]) {
cout << "RAHUL\n";
} else {
cout << "RAVI\n";
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram has a situation with him in which he needs to pick one name among Rahul and Ravi. So he decides to pick one name using a game. He asked Rahul and Ravi to play a game. The game is as follows.
First, Rahul throws dice and get a number X.
Then, Ravi throws dice and get a number Y.
Rahul wins the game if the sum on the dice is a prime number, and Ravi wins otherwise.
Given X and Y determines who wines the game.The first line of input will contain a single integer T, denoting the number of test cases.
The first and only line of each test case contains two space- separated integers X and Y.
<b>Constraints</b>
1 <b>≤</b> T <b>≤</b> 1000
1 <b>≤</b> X, Y <b>≤</b> 6For each test case, output on a new line the winner of the game: Rahul or Ravi.
Each letter of the output is printed in uppercase, i. e RAHUL, RAVI.Sample Input
3
1 2
3 4
5 5
Sample Output
RAHUL
RAHUL
RAVI, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int x=sc.nextInt();
for(int i=0;i<x; i++){
int m=sc.nextInt();
int n =sc.nextInt();
if(m+n==2)
System.out.println("RAHUL");
else if(m+n==3)System.out.println("RAHUL");
else if(m+n==5)System.out.println("RAHUL");
else if(m+n==7)System.out.println("RAHUL");
else if(m+n==11)System.out.println("RAHUL");
else System.out.println("RAVI");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alex always skips math class. As a punishment, his teacher has given him an array of size n where every number except one is the power of x.
Being poor in mathematics, Alex has asked for your help to solve the problem. Can you help him to find the bad number which is not a power of x?
<b>NOTE:</b>
It is guaranteed that there always exists an answer.The first line of the input contains the integers n and x
The following line contains n integers describing the array a
<b>Constraints</b>
2 ≤ n ≤ 1000
1 ≤ x ≤ 20
1 &le a<sub>i</sub> ≤ 10<sup>8</sup>For each test case, output a single line containing the bad number.input
6 7
16807 343 50 823543 2401 5764801
output
50, I have written this Solution Code: y=input().split()
n=int(y[0])
x=int(y[1])
a=input().split()
a=list(map(int,a))
k=0
for i in a:
r=i
while r!=1:
r/=x
if int(r)!=r:
print(i)
k=1
break
if k==1:
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alex always skips math class. As a punishment, his teacher has given him an array of size n where every number except one is the power of x.
Being poor in mathematics, Alex has asked for your help to solve the problem. Can you help him to find the bad number which is not a power of x?
<b>NOTE:</b>
It is guaranteed that there always exists an answer.The first line of the input contains the integers n and x
The following line contains n integers describing the array a
<b>Constraints</b>
2 ≤ n ≤ 1000
1 ≤ x ≤ 20
1 &le a<sub>i</sub> ≤ 10<sup>8</sup>For each test case, output a single line containing the bad number.input
6 7
16807 343 50 823543 2401 5764801
output
50, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, x;
cin >> n >> x;
vector<long long> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
vector<long long> powers;
long long MX = 1e8;
powers.push_back(1); // to store all the powers of x which are <= 10^8
while (true) {
long long curr = powers.back() * x;
if (curr <= MX) powers.push_back(curr);
else break;
}
for (int i = 0; i < n; i++) {
// checking if a[i] is in the powers array, i.e whether a[i] is a power of x or not
bool found = binary_search(powers.begin(), powers.end(), a[i]);
if (!found) {
cout << a[i];
return 0;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alex always skips math class. As a punishment, his teacher has given him an array of size n where every number except one is the power of x.
Being poor in mathematics, Alex has asked for your help to solve the problem. Can you help him to find the bad number which is not a power of x?
<b>NOTE:</b>
It is guaranteed that there always exists an answer.The first line of the input contains the integers n and x
The following line contains n integers describing the array a
<b>Constraints</b>
2 ≤ n ≤ 1000
1 ≤ x ≤ 20
1 &le a<sub>i</sub> ≤ 10<sup>8</sup>For each test case, output a single line containing the bad number.input
6 7
16807 343 50 823543 2401 5764801
output
50, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int x = sc.nextInt();
int ans = 0;
for(int i=0; i<n; i++){
int a = sc.nextInt();
if(!solve(a,x)){
System.out.print(a);
break;
}
}
}
static boolean solve(int a, int b){
while(a>1){
if(a%b!=0)
return false;
a = a/b;
}
return true;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer number N. The task is to generate all the positive binary numbers of N digits. These binary numbers should be in ascending order.The first line contains integer N - the number of digits in binary string
Constraints:
1 <= N <= 20Print all the positive binary numbers of N digits.Sample input 1:
1
Output:
1
Sample input 2:
2
Output:
10
11
Explanation:
2 can be written as 10 in binary
3 can be written as 11 in binary, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for (int i = (1 << (n - 1)); i < (1 << n); i++) {
System.out.println(Integer.toBinaryString(i));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer number N. The task is to generate all the positive binary numbers of N digits. These binary numbers should be in ascending order.The first line contains integer N - the number of digits in binary string
Constraints:
1 <= N <= 20Print all the positive binary numbers of N digits.Sample input 1:
1
Output:
1
Sample input 2:
2
Output:
10
11
Explanation:
2 can be written as 10 in binary
3 can be written as 11 in binary, I have written this Solution Code: n=int(input())
for i in range(1, 2**n):
a = bin(i)
a = a.replace('0b', '')
if len(a) == n:
print(a)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer number N. The task is to generate all the positive binary numbers of N digits. These binary numbers should be in ascending order.The first line contains integer N - the number of digits in binary string
Constraints:
1 <= N <= 20Print all the positive binary numbers of N digits.Sample input 1:
1
Output:
1
Sample input 2:
2
Output:
10
11
Explanation:
2 can be written as 10 in binary
3 can be written as 11 in binary, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define ll long long int
#define f(n) for(int i=0;i<n;i++)
#define fo(n) for(int j=0;j<n;j++)
#define foo(n) for(int i=1;i<=n;i++)
#define ff first
#define ss second
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define vp vector<pii>
#define test int tt; cin>>tt; while(tt--)
#define mod 1000000007
void fastio()
{
ios_base::sync_with_stdio(0); cin.tie(0);
#ifndef ONLINE_JUDGE
freopen("Input 1.txt", "r", stdin);
freopen("Output 1.txt", "w", stdout);
#endif
}
int main() {
fastio();
int n;
cin >> n;
for (int i = (1 << (n - 1)); i < (1 << n); i++) {
int x = i;
string s;
while (x) {
if (x & 1) {
s = '1' + s;
}
else s = '0' + s;
x /= 2;
}
cout << s << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are amazed at the relationship between John and Olivia. So you decided to tell that how many different pairs of relationships can be made from John and Olivia's relationship. For that, there is an array of length N and you need to divide it into K components such that each component contains an equal number of elements, no element is in two different components, and together it contains all the elements. You also need to divide in such a way that the parity of the sum of all the components is the same.
If there exists more than one K. Output the maximum such K.
But the array itself is not given only the number of odd, the number of even elements is given.The only line contains two integers A, B such that A denotes the number of even elements in the array and B denotes the number of odd elements in the array.
<b>Constraints</b>
1 <= A <= 10<sup>10</sup>
1 <= B <= 10<sup>10</sup>Output a single integer K the maximum value of the number of the components.Sample Input 1:
5 4
Sample Output 1:
3, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("popcnt")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
using cd = complex<double>;
const double PI = acos(-1);
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define fr first
#define sc second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifndef ONLINE_JUDGE
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
void solve(){
ll e, o; cin >> e >> o;
ll n = e+o;
ll ans = 1;
// a - even, b - odd
for(ll i = 1;i*i<=n;i++){
if(n%i != 0)continue;
ll x = i;
if(o%2 == 0){
// all x components is even
ll temp1 = 1;
if(2*x*((n/x)/2) < o)temp1 = 0;
// all x components is odd
ll temp2 = 1;
if(x%2 == 1)temp2 = 0;
if(x > o)temp2 = 0;
if(2*x*((n/x - 1)/2) < o - x)temp2 = 0;
if(temp1||temp2)ans = max(ans,x);
}
else{
// all x components is odd
ll temp = 1;
if(x%2 == 0)temp = 0;
if(x>o)temp = 0;
if(2*x*((n/x-1)/2) < o-x)temp = 0;
if(temp)ans = max(ans,x);
}
x = n/i;
if(o%2 == 0){
// all x components is even
ll temp1 = 1;
if(2*x*((n/x)/2) < o)temp1 = 0;
// all x components is odd
ll temp2 = 1;
if(x%2 == 1)temp2 = 0;
if(x > o)temp2 = 0;
if(2*x*((n/x - 1)/2) < o - x)temp2 = 0;
if(temp1||temp2)ans = max(ans,x);
}
else{
// all x components is odd
ll temp = 1;
if(x%2 == 0)temp = 0;
if(x>o)temp = 0;
if(2*x*((n/x-1)/2) < o-x)temp = 0;
if(temp)ans = max(ans,x);
}
}
cout << ans << "\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ll t = 1;
// cin >> t;
while(t--){
solve();
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code: public static long SumOfDivisors(long N){
long sum=0;
long c=(long)Math.sqrt(N);
for(long i=1;i<=c;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){sum+=N/i;}
}
}
return sum;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
def SumOfDivisors(num) :
# Final result of summation of divisors
result = 0
# find all divisors which divides 'num'
i = 1
while i<= (math.sqrt(num)) :
# if 'i' is divisor of 'num'
if (num % i == 0) :
# if both divisors are same then
# add it only once else add both
if (i == (num / i)) :
result = result + i;
else :
result = result + (i + num/i);
i = i + 1
# Add 1 to the result as 1 is also
# a divisor
return (result); , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] and a range [a, b]. The task is to partition the array around the range such that array is divided into three parts.
1) All elements smaller than a come first.
2) All elements in range a to b come next.
3) All elements greater than b appear in the end.
The individual elements of three sets can appear in any order. You are required to return the modified arranged array.
<b>Note:-</b>
In the case of custom input, you will get 1 if your code is correct else get a 0.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>threeWayPartition()</b> which contains following arguments.
A: input array list
low: starting integer of range
high: ending integer of range
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10<sup>4</sup>
1 <= A[i] <= 10<sup>5</sup>
1 <= low <= high <= 10<sup>5</sup>
The Sum of N over all test case doesn't exceed 10^5For each test case return the modified array.Sample Input:
2
5
1 8 3 3 4
3 5
3
1 2 3
1 3
Sample Output:
1 3 3 4 8
1 2 3
<b>Explanation:</b>
Testcase 1: First, the array has elements less than or equal to 3. Then, elements between 3 and 5. And, finally elements greater than 5. So, one of the possible outputs is 1 3 3 4 8.
Testcase 2: First, the array has elements less than or equal to 1. Then, elements between 1 and 3. And, finally elements greater than 3. So, the output is 1 2 3., I have written this Solution Code:
public static ArrayList<Integer> threeWayPartition(ArrayList<Integer> A, int lowVal, int highVal)
{
int n = A.size();
ArrayList<Integer> arr = A;
int start = 0, end = n-1;
for (int i=0; i<=end;)
{
// swapping the element with those at start
// if array element is less than lowVal
if (arr.get(i) < lowVal){
int temp=arr.get(i);
arr.add(i,arr.get(start));
arr.remove(i+1);
arr.add(start,temp);
arr.remove(start+1);
i++;
start++;
}
// swapping the element with those at end
// if array element is greater than highVal
else if (arr.get(i) > highVal){
int temp=arr.get(i);
arr.add(i,arr.get(end));
arr.remove(i+1);
arr.add(end,temp);
arr.remove(end+1);
end--;
}
// else just move ahead
else
i++;
}
return arr;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] and a range [a, b]. The task is to partition the array around the range such that array is divided into three parts.
1) All elements smaller than a come first.
2) All elements in range a to b come next.
3) All elements greater than b appear in the end.
The individual elements of three sets can appear in any order. You are required to return the modified arranged array.
<b>Note:-</b>
In the case of custom input, you will get 1 if your code is correct else get a 0.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>threeWayPartition()</b> which contains following arguments.
A: input array list
low: starting integer of range
high: ending integer of range
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10<sup>4</sup>
1 <= A[i] <= 10<sup>5</sup>
1 <= low <= high <= 10<sup>5</sup>
The Sum of N over all test case doesn't exceed 10^5For each test case return the modified array.Sample Input:
2
5
1 8 3 3 4
3 5
3
1 2 3
1 3
Sample Output:
1 3 3 4 8
1 2 3
<b>Explanation:</b>
Testcase 1: First, the array has elements less than or equal to 3. Then, elements between 3 and 5. And, finally elements greater than 5. So, one of the possible outputs is 1 3 3 4 8.
Testcase 2: First, the array has elements less than or equal to 1. Then, elements between 1 and 3. And, finally elements greater than 3. So, the output is 1 2 3., I have written this Solution Code: def threewayPartition(arr,low,high):
i=low-1
pivot = arr[high]
for j in range(low,high):
if arr[j]<=pivot:
i+=1
arr[i],arr[j]=arr[j],arr[i]
i+=1
arr[i],arr[high]=arr[high],arr[i]
return i
def quickSort(arr,low,high):
if low<high:
key=threewayPartition(arr,low,high)
quickSort(arr,low,key-1)
quickSort(arr,key+1,high)
return arr
t= int(input())
while t>0:
s = int(input())
arr = list(map(int,input().split()))
a,b = list(map(int,input().split()))
print(1)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings X and Y containing both uppercase and lowercase alphabets. The task is to find the length of the longest common substring.Input consist of two lines containing the strings X and Y respectively.
1 <= length(X), length(Y) <= 100Print in a single line the length of the longest common substring of the two strings.Input
abcdgh
acdghr
Output
4
Example:
Testcase 1: cdgh is the longest substring present in both of the strings., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str=br.readLine();
String str2=br.readLine();
int max_len=0;
for(int i=0;i<str.length();i++)
{
for(int j=0;j<str2.length();j++)
{
int len=0;
int u=i;
while(i<str.length() && j<str2.length() && str.charAt(i)==str2.charAt(j))
{
len++;
i++;
j++;
}
i=u;
if(max_len<len)
max_len=len;
}
}
System.out.println(max_len);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings X and Y containing both uppercase and lowercase alphabets. The task is to find the length of the longest common substring.Input consist of two lines containing the strings X and Y respectively.
1 <= length(X), length(Y) <= 100Print in a single line the length of the longest common substring of the two strings.Input
abcdgh
acdghr
Output
4
Example:
Testcase 1: cdgh is the longest substring present in both of the strings., I have written this Solution Code: /* Dynamic Programming solution to find length of the
longest common substring */
#include<iostream>
#include<string.h>
using namespace std;
int LCSubStr(string X, string Y, int m, int n)
{
// common suffix of X[0..i-1] and Y[0..j-1].
int LCSuff[m+1][n+1];
int result = 0; // To store length of the
// longest common substring
/* Following steps build LCSuff[m+1][n+1] in
bottom up fashion. */
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
// The first row and first column
// entries have no logical meaning,
// they are used only for simplicity
// of program
if (i == 0 || j == 0)
LCSuff[i][j] = 0;
else if (X[i-1] == Y[j-1])
{
LCSuff[i][j] = LCSuff[i-1][j-1] + 1;
result = max(result, LCSuff[i][j]);
}
else LCSuff[i][j] = 0;
}
}
return result;
}
/* Driver program to test above function */
int main()
{
string X, Y;
cin >> X >> Y;
int m = X.length();
int n = Y.length();
cout << LCSubStr(X, Y, m, n);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b>, of length <b>N</b> containing zero , positive and negative integers.
You need to find the length of the largest subarray whose sum of elements is <b>0</b>.The first line contains N denoting the size of the array A. Then in the next line contains N space-separated values of the array A.
<b>Constraints:-</b>
1 <= N <= 1e5
-1e6 <= A[i] <= 1e6Print the length of the largest subarray which has sum 0, If no subarray exist print -1.Sample Input:-
8
15 -2 2 -8 1 7 10 23
Sample Output:-
5
Explanation:-
-2 2 -8 1 7 is the required subarray
Sample Input:-
5
1 2 1 2 3
Sample Output:-
-1, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
public class Main {
public static void main(String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
int size = Integer.parseInt(line);
String str = br.readLine();
String[] strArray = str.split(" ");
int[] array = new int[size];
for (int i = -0; i < size; i++) {
array[i] = Integer.parseInt(strArray[i]);
}
int count = largestSubarray(array,size);
System.out.println(count);
}
static int largestSubarray(int[] array,int size){
int count = -1;
int sum = 0;
Map<Integer,Integer> mymap = new HashMap<>();
mymap.put(0,-1);
for(int i=0; i<array.length; i++){
sum += array[i];
if(mymap.containsKey(sum)){
count = Math.max(count, i-mymap.get(sum));
}
else{
mymap.put(sum,i);
}
}
if(count > 0){
return count;
}
else{
return -1;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b>, of length <b>N</b> containing zero , positive and negative integers.
You need to find the length of the largest subarray whose sum of elements is <b>0</b>.The first line contains N denoting the size of the array A. Then in the next line contains N space-separated values of the array A.
<b>Constraints:-</b>
1 <= N <= 1e5
-1e6 <= A[i] <= 1e6Print the length of the largest subarray which has sum 0, If no subarray exist print -1.Sample Input:-
8
15 -2 2 -8 1 7 10 23
Sample Output:-
5
Explanation:-
-2 2 -8 1 7 is the required subarray
Sample Input:-
5
1 2 1 2 3
Sample Output:-
-1, I have written this Solution Code: def maxLen(arr,n,k):
mydict = dict()
sum = 0
maxLen = 0
for i in range(n):
sum += arr[i]
if (sum == k):
maxLen = i + 1
elif (sum - k) in mydict:
maxLen = max(maxLen, i - mydict[sum - k])
if sum not in mydict:
mydict[sum] = i
return maxLen
n=int(input())
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
max_len=maxLen(arr,n,0)
if(max_len==0):
print ("-1")
else:
print (max_len), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b>, of length <b>N</b> containing zero , positive and negative integers.
You need to find the length of the largest subarray whose sum of elements is <b>0</b>.The first line contains N denoting the size of the array A. Then in the next line contains N space-separated values of the array A.
<b>Constraints:-</b>
1 <= N <= 1e5
-1e6 <= A[i] <= 1e6Print the length of the largest subarray which has sum 0, If no subarray exist print -1.Sample Input:-
8
15 -2 2 -8 1 7 10 23
Sample Output:-
5
Explanation:-
-2 2 -8 1 7 is the required subarray
Sample Input:-
5
1 2 1 2 3
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unordered_map<long long,int> m;
int n,k;
cin>>n;
long a[n];
int ans=-1;
for(int i=0;i<n;i++){cin>>a[i];if(a[i]==0){ans=1;}}
long long sum=0;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==0){ans=max(i+1,ans);}
if(m.find(sum)==m.end()){m[sum]=i;}
else{
ans=max(i-m[sum],ans);
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers and another integer D, your task is to perform D right circular rotations on the array and print the modified array.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>circularRotate() </b> which takes the deque, number of rotation and the number of elements in deque as parameters.
Constraint:
1 <= T <= 100
1 <= N <= 100000
1 <= elements <= 10000
1 <= D <= 100000You don't need to print anything you just need to complete the function.Input:
2
5
1 2 3 4 5
6
4
1 2 3 4
7
Output:
5 1 2 3 4
2 3 4 1, I have written this Solution Code: static void circularRotate(Deque<Integer> deq, int d, int n)
{
// Push first d elements
// from last to the beginning
for (int i = 0; i < d%n; i++) {
int val = deq.peekLast();
deq.pollLast();
deq.addFirst(val);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code:
import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args)throws Exception{ new Main().run();}
long mod=1000000000+7;
long tsa=Long.MAX_VALUE;
void solve() throws Exception
{
long X=nl();
div(X);
out.println(tsa);
}
long cal(long a,long b, long c)
{
return 2l*(a*b + b*c + c*a);
}
void div(long n)
{
for (long i=1; i*i<=n; i++)
{
if (n%i==0)
{
if (n/i == i)
{
all_div(i, i);
}
else
{
all_div(i, n/i);
all_div(n/i, i);
}
}
}
}
void all_div(long n , long alag)
{
ArrayList<Long> al = new ArrayList<>();
for (long i=1; i*i<=n; i++)
{
if (n%i==0)
{
if (n/i == i)
tsa=min(tsa,cal(i,i,alag));
else
{
tsa=min(tsa,cal(i,n/i,alag));
}
}
}
}
private byte[] buf=new byte[1024];
private int index;
private InputStream in;
private int total;
private SpaceCharFilter filter;
PrintWriter out;
int min(int... ar){int min=Integer.MAX_VALUE;for(int i:ar)min=Math.min(min, i);return min;}
long min(long... ar){long min=Long.MAX_VALUE;for(long i:ar)min=Math.min(min, i);return min;}
int max(int... ar) {int max=Integer.MIN_VALUE;for(int i:ar)max=Math.max(max, i);return max;}
long max(long... ar) {long max=Long.MIN_VALUE;for(long i:ar)max=Math.max(max, i);return max;}
void reverse(int a[]){for(int i=0;i<a.length>>1;i++){int tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
void reverse(long a[]){for(int i=0;i<a.length>>1;i++){long tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
String reverse(String s){StringBuilder sb=new StringBuilder(s);sb.reverse();return sb.toString();}
void shuffle(int a[]) {
ArrayList<Integer> al = new ArrayList<>();
for(int i=0;i<a.length;i++)
al.add(a[i]);
Collections.sort(al);
for(int i=0;i<a.length;i++)
a[i]=al.get(i);
}
long lcm(long a,long b)
{
return (a*b)/(gcd(a,b));
}
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long expo(long p,long q)
{
long z = 1;
while (q>0) {
if (q%2 == 1) {
z = (z * p)%mod;
}
p = (p*p)%mod;
q >>= 1;
}
return z;
}
void run()throws Exception
{
in=System.in; out = new PrintWriter(System.out);
solve();
out.flush();
}
private int scan()throws IOException
{
if(total<0)
throw new InputMismatchException();
if(index>=total)
{
index=0;
total=in.read(buf);
if(total<=0)
return -1;
}
return buf[index++];
}
private int ni() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = scan();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = scan();
} while (!isSpaceChar(c));
return res * sgn;
}
private long nl() throws IOException
{
long num = 0;
int b;
boolean minus = false;
while ((b = scan()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = scan();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = scan();
}
}
private double nd() throws IOException{
return Double.parseDouble(ns());
}
private String ns() throws IOException {
int c = scan();
while (isSpaceChar(c))
c = scan();
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c))
res.appendCodePoint(c);
c = scan();
} while (!isSpaceChar(c));
return res.toString();
}
private String nss() throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
return br.readLine();
}
private char nc() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
return (char) c;
}
private boolean isWhiteSpace(int n)
{
if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
return true;
return false;
}
private boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return isWhiteSpace(c);
}
private interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code: m =int(input())
def print_factors(x):
factors=[]
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return(factors)
area=[]
factors=print_factors(m)
for a in factors:
for b in factors:
for c in factors:
if(a*b*c==m):
area.append((2*a*b)+(2*b*c)+(2*a*c))
print(min(area)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int x;
cin >> x;
int ans = 6*x*x;
for (int i=1;i*i*i<=x;++i)
if (x%i==0)
for (int j=i;j*j<=x/i;++j)
if (x/i % j==0) {
int k=x/i/j;
int cur=0;
cur=i*j+i*k+k*j;
cur*=2;
ans=min(ans,cur);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array A of n integers, you need to rearrange the array to the lexicographically smallest array such that the elements of array appear in the following manner:
A[0] >= A[1] <= A[2] >= A[3] <= A[4] ....The first line of input contains an integer n, the number of elements in array.
The next line contains n integers, the elements of the array.
Constraints
1 <= n <= 100000
1 <= A[i] <= 1000000000Output the required array with space separated integers.Sample Input
4
3 2 1 4
Sample Output
2 1 4 3
Explanation
Other possible array satisfying the condition is [4 1 3 2] but that is not lexicographically smallest., I have written this Solution Code: n = int(input())
nums = list(map(int,input().split()))
nums.sort()
for i in range(0,n-1,2):
print(nums[i+1],nums[i],end=" ")
if n%2 != 0:
print(nums[n-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array A of n integers, you need to rearrange the array to the lexicographically smallest array such that the elements of array appear in the following manner:
A[0] >= A[1] <= A[2] >= A[3] <= A[4] ....The first line of input contains an integer n, the number of elements in array.
The next line contains n integers, the elements of the array.
Constraints
1 <= n <= 100000
1 <= A[i] <= 1000000000Output the required array with space separated integers.Sample Input
4
3 2 1 4
Sample Output
2 1 4 3
Explanation
Other possible array satisfying the condition is [4 1 3 2] but that is not lexicographically smallest., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine().trim());
int arr[] = new int[n];
String inputLine[] = br.readLine().trim().split(" ");
for(int i=0; i<n; i++){
arr[i]=Integer.parseInt(inputLine[i]);
}
Strange obj = new Strange();
obj.convert(arr, n);
StringBuffer sb = new StringBuffer();
for (int i = 0; i < n; i++)
sb.append(arr[i] + " ");
System.out.println(sb);
}
}
class Strange{
public static void convert(int arr[], int n){
Arrays.sort(arr);
for(int i=0;i<=n-2;i+=2){
int temp=arr[i];
arr[i]=arr[i+1];
arr[i+1]=temp;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array A of n integers, you need to rearrange the array to the lexicographically smallest array such that the elements of array appear in the following manner:
A[0] >= A[1] <= A[2] >= A[3] <= A[4] ....The first line of input contains an integer n, the number of elements in array.
The next line contains n integers, the elements of the array.
Constraints
1 <= n <= 100000
1 <= A[i] <= 1000000000Output the required array with space separated integers.Sample Input
4
3 2 1 4
Sample Output
2 1 4 3
Explanation
Other possible array satisfying the condition is [4 1 3 2] but that is not lexicographically smallest., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int n;
cin>>n;
vector<int> A;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
A.pu(a);
}
sort(A.begin(),A.end());
for(int i=0;i<n;i++)
{
i++;
if(i==n) cout<<A[i-1]<<" ";
else{
cout<<A[i]<<" "<<A[i-1]<<" ";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
Reader sc=new Reader();
int n=sc.nextInt();
int b=sc.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int j, max;
long mul=1;
Deque<Integer> dq= new LinkedList<Integer>();
for (int i = 0;i<b;i++)
{
while (!dq.isEmpty() && a[i] >=a[dq.peekLast()])
dq.removeLast();
dq.addLast(i);
}
mul=((mul%1000000007)*(a[dq.peek()]%1000000007))%1000000007;
for (int i=b; i < n;i++)
{
while ((!dq.isEmpty()) && dq.peek() <=i-b)
dq.removeFirst();
while ((!dq.isEmpty()) && a[i]>=a[dq.peekLast()])
dq.removeLast();
dq.addLast(i);
mul=((mul%1000000007)*(a[dq.peek()]%1000000007))%1000000007;
}
System.out.println(mul%1000000007);
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: from collections import deque
deq=deque()
n,k=list(map(int,input().split()))
array=list(map(int,input().split()))
for i in range(k):
while(deq and array[deq[-1]]<=array[i]):
deq.pop()
deq.append(i)
ans=1
for j in range(k,n):
ans=ans*array[deq[0]]
ans=(ans)%1000000007
while(deq and deq[0]<=j-k):
deq.popleft()
while(deq and array[deq[-1]]<=array[j]):
deq.pop()
deq.append(j)
ans=ans*array[deq[0]]
ans=(ans)%1000000007
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
deque<int> q;
for(int i = 1; i <= k; i++){
while(!q.empty() && a[q.back()] <= a[i])
q.pop_back();
q.push_back(i);
}
int ans = a[q.front()];
for(int i = k+1; i <= n; i++){
if(q.front() == i-k)
q.pop_front();
while(!q.empty() && a[q.back()] <= a[i])
q.pop_back();
q.push_back(i);
ans = (ans*a[q.front()]) % mod;
}
cout << ans;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int temp;
for(int i=1;i<n;i++){
if(a[i]<a[i-1]){
for(int j=i;j>0;j--){
if(a[j]<a[j-1]){
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
else{
break;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: def bubbleSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from a range he selected, but somehow he lost one element from the array (other the first and the last element).
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space- separated integers.
Constraints:-
1 < = N < = 100000
1 < = elements < = 100000
Note:- It is guaranteed that the maximum element - minimum element is equal to N-1 and all the elements will be different.Print the missing integer.Sample Input:-
3
3 5
Sample Output:-
4
Sample Input:-
5
9 13 12 11
Sample Output:-
10, I have written this Solution Code: N=int(input())
arr_of_int=[None] * (N-1)
arr_of_int=list(map(int, input().split()))
min_elem=min(arr_of_int)
max_elem=max(arr_of_int)
arrsum=0
for i in range(len(arr_of_int)):
arrsum+=arr_of_int[i]
sum_of_arr=0
i=min_elem
while i<=max_elem:
sum_of_arr+=i
i+=1
print((sum_of_arr-arrsum)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from a range he selected, but somehow he lost one element from the array (other the first and the last element).
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space- separated integers.
Constraints:-
1 < = N < = 100000
1 < = elements < = 100000
Note:- It is guaranteed that the maximum element - minimum element is equal to N-1 and all the elements will be different.Print the missing integer.Sample Input:-
3
3 5
Sample Output:-
4
Sample Input:-
5
9 13 12 11
Sample Output:-
10, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
long a[] = new long[m];
long sum=0;
long x = 10000000;
long y=0;
for(int i=0;i<m-1;i++){
a[i]= sc.nextInt();
if(a[i]<x){
x=a[i];
}
if(a[i]>y){
y=a[i];
}
sum+=a[i];
}
long ans1 = x*(x-1);
ans1/=2;
long ans2 = y*(y+1);
ans2/=2;
long res= ans2-ans1-sum;
System.out.print(res);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: m,n=map(int ,input().split())
matrix=[]
for i in range(m):
l1=[eval(x) for x in input().split()]
matrix.append(l1)
l2=[]
for coloumn in range(n):
sum1=0
for row in range(m):
sum1+= matrix[row][coloumn]
l2.append(sum1)
print(max(l2))
'''for row in range(n):
sum2=0
for col in range(m):
sum2 += matrix[row][col]
print(sum2)''', In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: // mat is the matrix/ 2d array
// the dimensions of array are:- a rows, b columns
function colMaxSum(mat,a,b) {
// write code here
// do not console.log
// return the answer as a number
let idx = -1;
// Variable to store max sum
let maxSum = Number.MIN_VALUE;
// Traverse matrix column wise
for (let i = 0; i < b; i++) {
let sum = 0;
// calculate sum of column
for (let j = 0; j < a; j++) {
sum += mat[j][i];
}
// Update maxSum if it is
// less than current sum
if (sum > maxSum) {
maxSum = sum;
// store index
idx = i;
}
}
let res;
res = [idx, maxSum];
// return result
return maxSum;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n,m;
cin>>n>>m;
int a[m];
for(int i=0;i<m;i++){
a[i]=0;
}
int x;
int sum=0;
FOR(i,n){
FOR(j,m){
cin>>x;
a[j]+=x;
sum=max(sum,a[j]);
}
}
out(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int a[][] = new int[m][n];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
a[i][j]=sc.nextInt();
}
}
int sum=0;
int ans=0;
for(int i=0;i<n;i++){
sum=0;
for(int j=0;j<m;j++){
sum+=a[j][i];
}
if(sum>ans){ans=sum;}
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers and another integer D, your task is to perform D right circular rotations on the array and print the modified array.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>circularRotate() </b> which takes the deque, number of rotation and the number of elements in deque as parameters.
Constraint:
1 <= T <= 100
1 <= N <= 100000
1 <= elements <= 10000
1 <= D <= 100000You don't need to print anything you just need to complete the function.Input:
2
5
1 2 3 4 5
6
4
1 2 3 4
7
Output:
5 1 2 3 4
2 3 4 1, I have written this Solution Code: static void circularRotate(Deque<Integer> deq, int d, int n)
{
// Push first d elements
// from last to the beginning
for (int i = 0; i < d%n; i++) {
int val = deq.peekLast();
deq.pollLast();
deq.addFirst(val);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a circular linked list consisting of N nodes and an integer K, your task is to add the integer K at the end of the list.
<b>Note:
Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Insertion()</b> that takes head node of circular linked list and the integer K as parameter.
Constraints:
1 <=N <= 1000
1 <= Node.data, K<= 1000Return the head node of the modified circular linked list.Sample Input 1:-
3
1- >2- >3
4
Sample Output 1:-
1- >2- >3- >4
Sample Input 2:-
3
1- >3- >2
1
Sample Output 2:-
1- >3- >2- >1, I have written this Solution Code: public static Node Insertion(Node head, int K){
Node node=head;
while ( node.next != head)
{node = node.next; }
Node temp = new Node(K);
node.next=temp;
temp.next=head;
return head;}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int[] arr=new int[n];
String[] s=br.readLine().trim().split(" ");
for(int i=0;i<arr.length;i++)
arr[i]=Integer.parseInt(s[i]);
int max=Integer.MIN_VALUE,c=0;
for(int i=0;i<arr.length;i++){
if(max==arr[i])
c++;
if(max<arr[i]){
max=arr[i];
c=1;
}
}
System.out.println(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: n=int(input())
a=list(map(int,input().split()))
print(a.count(max(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(101, 0);
For(i, 0, n){
int x; cin>>x;
a[x]++;
}
for(int i=100; i>=1; i--){
if(a[i]){
cout<<a[i];
return;
}
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Initialize two variables <code>name</code> and <code>age</code> with values of data types <code>string</code> and <code>number</code> respectively.
Create another variable <code>greet</code> with the value of "My name is <code>name</code> and I am <code>age</code> years old" and print the value of <code>greet</code> in the console.
Note: Generate Expected Output section will not work for this question
DO NOT CONSOLE.LOG the variables, just declare them.There is no input required for this questionThere is no output required for this questionIf value of <code>name</code> is Raj and value of <code>age</code> is 18, then <code>greet</code> should be storing "My name is Raj and I am 18 years old", I have written this Solution Code: const name = "John Doe";
const age = 30;
const greet = `My name is ${name} and I am ${age} years old`;
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void sieve(boolean prime[], int n) {
int i,j;
for(i = 0; i <= n; i++)
prime[i] = true;
for(i = 2; i*i <= n; i++)
if(prime[i])
for(j = i*i; j<=n; j+=i)
prime[j] = false;
}
public static void main (String[] args) throws IOException {
int num = 10000005;
boolean prime[] = new boolean[num+1];
sieve(prime, num);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine().trim());
while(T --> 0) {
int n = Integer.parseInt(br.readLine().trim());
int count = 0;
for(int i=(n/2)+1; i<=n; i++)
if(prime[i])
count++;
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
bool a[max1];
long b[max1];
void pre(){
b[0]=0;b[1]=0;
for(int i=0;i<max1;i++){
a[i]=false;
}
long cnt=0;
for(int i=2;i<max1;i++){
if(a[i]==false){
cnt++;
for(int j=i+i;j<=max1;j=j+i){a[j]=true;}
}
b[i]=cnt;
}
}
int main(){
pre();
int t;
cin>>t;
while(t--){
long n;
cin>>n;
cout<<(b[n]-b[(n)/2])<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
long cnt=0;
long ans=0;
for(int i=0;i<n;i++){
if(a[i]&1){cnt++;}
else{
ans+=(cnt*(cnt+1))/2;
cnt=0;
}
}
ans+=(cnt*(cnt+1))/2;
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
c=0
result=0
for i in arr:
if i % 2 != 0:
c += 1
else:
result += (c*(c+1)) / 2
c=0
result += (c*(c+1)) / 2
print(int(result)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader inputStreamReader=new InputStreamReader(System.in);
BufferedReader reader=new BufferedReader(inputStreamReader);
int n =Integer.parseInt(reader.readLine());
String str=reader.readLine();
String[] strarr=str.split(" ");
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(strarr[i]);
}
long noOfsubArrays=0;
int start=0;
boolean sFlag=false;
for(int i=0;i<n;i++){
if(arr[i]%2!=0){
if(!sFlag){
start=i;
noOfsubArrays++;
sFlag=true;
continue;
}
noOfsubArrays+=2;
int temp=start;
temp++;
while(temp<i){
temp++;
noOfsubArrays++;
}
}else if(arr[i]%2==0 && sFlag){
sFlag=false;
}
}
System.out.println(noOfsubArrays);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.